Question

Find the value of \(\int\int_0^y \frac{2xy^5}{\sqrt{1+x^2 y^2-y^4}} dxdy\).

(a) \(2[\frac{y^4}{4}- \frac{2}{3} (1-y^4)^{\frac{3}{2}}]\)

(b) \(2[\frac{y^4}{4}- (1-y^4)^{\frac{3}{2}}]\)

(c) \(2[\frac{y^4}{4}-\frac{2}{3} (1-y^4)^{\frac{3}{2}}]\)

(d) \(2[\frac{y^3}{3}-\frac{2}{3} (1-y^4)^{\frac{3}{2}}]\)

I had been asked this question in my homework.

Query is from Double Integrals in section Multiple Integrals of Engineering Mathematics

Answer 1

Right option is (c) \(2[\frac{y^4}{4}-\frac{2}{3} (1-y^4)^{\frac{3}{2}}]\)

To explain: Given, f(x)=\(\int\int_0^y \frac{2xy^5}{\sqrt{1+x^2 y^2-y^4}} dxdy\)

=\(\int\int_0^y \frac{1}{y} \frac{2xy^5}{\sqrt{(\frac{1-y^4}{y^2})+x^2}} dxdy=\int 2y^4 \left |(\frac{1-y^4}{y^2})+x^2\right |_0^y dy\)

\(=2\int [y^3-\sqrt{1-y^4}y^3]dy=2[\frac{y^4}{4}-\frac{2}{3} (1-y^4)^{3/2}]\)